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If y = \frac {x^2 - y^2}{x-y} ( x \ne y ), then what is the [#permalink]
 16 Oct 2011, 04:07
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If y = \frac {x^2 - y^2}{x-y} ( x \ne y ), then what is the value of y ? 1. x + y = 3 2. x - y = -3 (C) 2008 GMAT Club - m02#25Hi All- I'm struggling to see the logic behind the OA in this question. Taking the information from the question stem, we get y= (x+y)(x-y)/ (x-y), therefore y=x+y. Its easy to see that statement 1 is sufficient, but for statement 2, we will still get y=x+y because y=-3(x+y)/-3. We can pick a number of different values for x-y that equal -3 that fit the question stem (x is not equal to y). Can someone with better math skills than myself explain why statement 2 is sufficient? Cheers! -Ken
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Re: Tricky Algebra [#permalink]
16 Oct 2011, 04:14
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kenman wrote: If y = \frac {x^2 - y^2}{x-y} ( x \ne y ), then what is the value of y ?
1. x + y = 3
2. x - y = -3
(C) 2008 GMAT Club - m02#25
Hi All-
I'm struggling to see the logic behind the OA in this question. Taking the information from the question stem, we get y= (x+y)(x-y)/ (x-y), therefore y=x+y. Its easy to see that statement 1 is sufficient, but for statement 2, we will still get y=x+y because y=-3(x+y)/-3. We can pick a number of different values for x-y that equal -3 that fit the question stem (x is not equal to y). Can someone with better math skills than myself explain why statement 2 is sufficient? Cheers!
-Ken Mark my words bro, if you follow the way you are solving the questions you will face a difficult time with DS questions. Always simply the equations till end, else you will complicate the things for you... - GOLDEN RULE. now when you got y =x+y => x must be 0 now when you will put x=0 in both the statements you will have unique value of Y. => both are sufficient. Now spend at least 15-20 mins on the advice I gave you and try to put in every question. Always simplify and then go ahead.
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Re: Tricky Algebra [#permalink]
16 Oct 2011, 12:44
even i mistakenly chose A...But its little difficult to understand why D?
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Re: Tricky Algebra [#permalink]
16 Oct 2011, 13:13
prateekbhatt wrote: even i mistakenly chose A...But its little difficult to understand why D? {x^2 - y^2} = (x+y)*(x-y)The original expression ends up being: y = x + y thus x=0. From here, both (1) and (2) can be solved: (1) y=3(2) -y=-3; y=3
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Re: Tricky Algebra [#permalink]
16 Oct 2011, 14:06
LGOdream wrote: prateekbhatt wrote: even i mistakenly chose A...But its little difficult to understand why D? {x^2 - y^2} = (x+y)*(x-y)The original expression ends up being: y = x + y thus x=0. From here, both (1) and (2) can be solved: (1) y=3(2) -y=-3; y=3Ohh got it.....I was thinking that if statement 1 is solving the purpose then why to check statement 2 ....i was wrong... Thanks for solving.....
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Re: Tricky Algebra [#permalink]
22 Oct 2011, 12:13
+1 for D.
The given input is a good indication of what the questions expects (x not equal to y).
Simplifying the given equations we get:
x(x-y) = 0, since x -y = 0 or x is not equal to y, we know that x = 0.
Substituting x=0 in both conditions will give us the value of y. Hence D.
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Re: If y = \frac {x^2 - y^2}{x-y} ( x \ne y ), then what is the [#permalink]
16 Dec 2011, 07:07
Excellent question kenman. Definitely worth another look. Its easy to miss out on the minor details and make careless mistakes as I did. D is the right answer. Thanks for the explanation gurpreetsingh. Kudos to both..
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Re: If y = \frac {x^2 - y^2}{x-y} ( x \ne y ), then what is the [#permalink]
03 Jan 2012, 22:45
D. From question stem we know y = (x + y) (x-y) / (x-y) y = (x + y) , y = ? (1) x+ y = 3. so y = 3 SUFF. (2) x -y = -3 x = -3 + y Subs value of x in rephrased question stem y = -3 + y + y y = 3. SUFF.
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Re: If y = \frac {x^2 - y^2}{x-y} ( x \ne y ), then what is the
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03 Jan 2012, 22:45
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